Engineering Dynamics
ENGR 2090
Section 02 M R 10:00AM – 11:50AM
Lecture 10
PROFESSOR TAYLOR
TAYLOK5@RPI.EDU
OFFICE HOURS:
JEC 2032 – TUESDAYS 1:00 PM TO 3:00 PM
Angular Impulse and Momentum
H is the symbol we will use to represent ‘Angular Momentum’ (a physics course might
use L). Angular Momentum is the tendency of an object to keep spinning.
𝑁
𝑒3
𝐻
𝑖
𝑜
≡ 𝑟 𝑜𝑖 × 𝑚𝑖 𝑁𝑉 𝑖
‘N’
𝑖
𝑟 𝑜𝑖
𝑒2
𝑒1
Temporal Derivative of Angular
Momentum
Now, let’s look at how the time derivative of this equation behaves. (In other words,
we want to be able to determine how moments cause a change in angular momentum.
That is the destination).
𝑁
𝑑
𝑑𝑡
𝑁
𝐻
𝑖
𝑁
𝑜
𝑑 𝑜𝑖
=
𝑟 × 𝑚𝑖 𝑁 𝑉 𝑖
𝑑𝑡
This is a cross product. The day has come when the distinction is
crazy important
Note, we are going to use an identity : 𝐴 × 𝑐𝐵 = 𝑐 𝐴 × 𝐵
Distributing a derivative
This is how this temporal derivative operator interacts with the cross product:
𝑁
𝑑
𝑑𝑡
𝑁
𝐻
𝑖
𝑜
= 𝑚𝑖
𝑁 𝑖
𝑑𝑟 𝑜𝑖 𝑁 𝑖
𝑑
𝑉
𝑜𝑖
× 𝑉 + 𝑟 ×
𝑑𝑡
𝑑𝑡
That expression is formatted oddly. Below is the exact same expression, but I altered
the formatting. (this is not a math issue, this is a PowerPoint issue)
𝑁
𝑑
𝑑𝑡
𝑁𝐻 𝑖 𝑜
= 𝑚𝑖
𝑑 𝑜𝑖 𝑁 𝑖
𝑑 𝑁 𝑖
𝑜𝑖
𝑟 × 𝑉 + 𝑟 ×
𝑉
𝑑𝑡
𝑑𝑡
Derivatives of position and velocity
The derivative of position is velocity. Let’s use the definitions of velocity and
acceleration:
𝑑 𝑜𝑖 𝑁 𝑖
𝑟 = 𝑉
𝑑𝑡
𝑑 𝑁 𝑖 𝑁 𝑖
𝑉 = 𝑎
𝑑𝑡
𝑜
Let’s rewrite the last equation from the previous slide then:
𝑁
𝑑
𝑑𝑡
𝑁𝐻 𝑖 𝑜
= 𝑚𝑖
𝑁𝑉 𝑖 𝑜
× 𝑁𝑉 𝑖 + 𝑟 𝑜𝑖 × 𝑁𝑎𝑖
Review of the cross product
What is this expression equal to:
𝑁 𝑖 𝑜
𝑉
× 𝑁𝑉 𝑖
Oops: I forgot. I use shorthand when a velocity is relative to an origin
𝑁 𝑖
𝑉 × 𝑁𝑉 𝑖
Yep, this term goes to zero.
And we are getting close to something
useful
𝑁
𝑑
𝑑𝑡
𝑁
𝐻
𝑖
𝑜
= 𝑚𝑖 𝑟 𝑜𝑖 × 𝑁𝑎𝑖
Note, we are going to use an identity : 𝐴 × 𝑐𝐵 = 𝑐 𝐴 × 𝐵
𝑁
𝑑
𝑑𝑡
𝑁
𝐻
𝑖
𝑜
= 𝑟 𝑜𝑖 × 𝑚𝑖 𝑁𝑎𝑖
Discussion : What is that last term?
Differential form
𝑁
𝑑
𝑑𝑡
𝑁
𝑁
𝑖
𝑑 𝐻
𝐻
𝑜
𝑖
𝑜
=𝑟
𝑁
= 𝑀
𝑜𝑖
𝑖
𝑁 𝑖
𝑁
× 𝑚𝑖 𝑎 = 𝑀
𝑖
𝑜 𝑑𝑡
This expression can be integrated, to find:
𝑡2
𝑡1
𝑁
𝑀
𝑖
𝑜 𝑑𝑡
𝑖
𝑜
𝑖
= 𝐻2 − 𝐻1 𝑜
𝑁
𝑁
𝑜
Ok, but what if no moment couples are
applied?
If no external moment couples or if no external forces are applied to the system:
𝑡2 𝜂
𝑡1 𝑖=1
𝜂
𝑁
𝑀
𝑖
𝑜 𝑑𝑡
𝑁
=0=
𝑖
𝑖
𝐻2 − 𝐻1 𝑜
𝑖=1
This statement is the Conservation of Angular Momentum
𝑜
𝑁
Collisions
Impact – Not all impacts are equal. Sometimes energy is lost to heat or permanent
deformation. What happens when we lose energy?
The first ‘state’ we call 1 is before collision
A
B
𝑁 𝐴
𝑉1
> 𝑁𝑉1𝐵
State II - collision
During collision, we can treat these as one particle. There is a velocity associated with
the system
A
B
State III – After Collision
A
B
𝑁 𝐴
𝑉2
<
𝑁 𝐵
𝑉2
Basic Model – Coefficient of Restitution
During deformation, there should be an ‘intermediate’ velocity
A
𝑁 𝐴
𝑉1
𝑁 𝐵
𝑉1
B
𝐴
→ 𝑁𝑉𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
→ 𝑁𝑉2𝐴
→
𝑁 𝐵
𝑉𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛
→ 𝑁𝑉2𝐵
Impulse
Impulse is a force applied through time, and the result is a change in momentum. We
are going to look at two ‘forces’
1. One force is the force needed to deform the objects. Let’s call this 𝐹𝑑
2. One force is the force of the objects A and B returning to their original shapes.
It helps to imagine them as springs ‘bouncing’ back to their original shape.
Let’s call this 𝐹𝑟
We will look at this just for particle A first.
𝑒=
𝑡2
𝐹
𝑑𝑡
𝑟
𝑡𝑐
𝑡𝑐
𝐹
𝑑𝑡
𝑑
𝑡1
𝑚𝐴 𝑣2𝐴 − 𝑚𝐴 𝑣𝑐𝐴
=
𝐴
𝐴
𝑚𝐴 𝑣𝑐 − 𝑚𝐴 𝑣1
Cancelling out the mass…
And this simplifies to two expressions:
𝑒=
𝑣2𝐴
𝐴
𝑣𝑐
−
−
𝑣𝑐𝐴
𝐴
𝑣1
𝑒=
𝐵
𝑣2
𝐵
𝑣𝑐
− 𝑣𝑐𝐵
𝐵
− 𝑣1
The efficiencies are equal : At present, our model does not allow us to lose energy or
momentum to any other process.
Your takeaway
I will spare you the gory details of the algebra, but efficiency will boil down to:
𝑣2𝐵 − 𝑣2𝐴
𝑒= 𝐴
𝐵
𝑣1 − 𝑣1
Special note for during an exam… use the
version three slides further along
In this context, e is the efficiency, the ratio of the difference in the end velocities to the
difference in the start velocities.
ESTABLISH A REFERENCE FRAME. THIS IS ONLY TRUE ON THE LINE DEFINED BY THE
LINE CONNECTING THE CENTERS OF MASS OF THESE OBJECTS (exception : one of the
objects is ‘fixed’ and can’t change position.)
Schematic Example
Momentum is quasi conserved at some efficiency e in the 𝑛 direction for the system
Momentum is conserved in the 𝑡 direction for both particles. (there is no deformation in
this direction)
Mathematically Expressed
There is another way to express this mathematically
𝑉1𝐴
𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡
= 𝑉2𝐴
𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡
Or maybe we could do something like:
𝐴
𝐴
𝑣1𝑡
= 𝑣2𝑡
The tangential component of velocity of a particle during {a collision event of two
particles} is constant.
The efficiency in the normal direction
𝐵
𝐴
𝑣2𝑛
− 𝑣2𝑛
𝑒= 𝐴
𝐵
𝑣1𝑛 − 𝑣1𝑛
BUT THIS STATEMENT MEANS NOTHING IF
YOU DO NOT DRAW A BASIS, AND, ALIGN
ONE OF THE BASIS VECTORS WITH THE LINE
SEGMENT THAT CONNECTS THE CENTER OF
MASS OR THE CENTROID OF THE OBJECTS
COLLIDING.
Elastic and inelastic collisions
A perfectly elastic collision means that the efficiency is 1
An inelastic collision means the efficiency is zero. Literally, this is when two particles
come together and stick together. When a collision behaves in this way, this is an
inelastic collision.
Elastic Collisions
We will consider elastic Collisions to have two assumptions:
1. There are no net external forces to the system
2. There is no momentum lost during the collision to other ‘physics’ or ‘chemistry’
𝐴
𝑃1
−
𝐴
𝑃2
=
𝐵
𝑃2
−
𝐵
𝑃1
If we consider multiple bodies all making elastic collisions:
𝜂
𝑃2𝑖 − 𝑃1𝑖 = 0
𝑖=1
Lecture 10 Example 1
The assembly starts from rest and reaches an
angular speed of 150 rev/min under the action
of a 20-N force T applied to the string for t
seconds. Determine t. Neglect friction and all
masses except those the four 3-kg spheres,
which may be treated as particles.
Step 1 : Establish some frames of
reference
One of the
balls spinning
is called ‘i’
Note that
point O is
located at the
center of the
spinning 4
balls.
Step 2 : Governing equation.
So yes, I probably need the angular momentum equation:
𝑡2
𝑁
𝑀
𝑖
𝑜 𝑑𝑡
𝑡1
𝑁
𝑖
𝑜
𝑁
𝑖
= 𝐻2 − 𝐻1
𝑜
I forget, what is H again?
𝑁
𝐻
𝑖
𝑜
≡ 𝑟 𝑜𝑖 × 𝑚𝑖 𝑁𝑉 𝑖
Hmmm… looks like I will need an expression for velocity. Let’s start there.
Relative velocity
The velocity of particle I will have something to do with the angular momentum about
the line of rotation. Point O is on that line. Let’s try to express the velocity of a single
ball i with respect to point O
𝑁 𝑖
𝑁 𝑂
𝑉 = 𝑉 +
𝐴 𝑖 𝑂
𝑉
+ 𝑁𝜔 𝐴 × 𝑟 𝑂𝑖
Discussion : What do we do with these terms?
𝑁 𝑖
𝑁 𝑂
𝑉 = 𝑉 +
𝐴 𝑖 𝑂
𝑉
+ 𝑁𝜔 𝐴 × 𝑟 𝑂𝑖
Discussion : What do we do with these terms?
Simplifying
Zero. In frame A, is is not
moving toward or away
from O
𝑁 𝑖
𝑁 𝑂
𝑉 = 𝑉 +
zero. In frame N,
point O has no
translational
velocity.
𝐴 𝑖 𝑂
𝑉
+ 𝑁𝜔 𝐴 × 𝑟 𝑂𝑖
This represents the rate of
rotation of reference frame A
Bringing the governing equations
together
𝑁 𝑖
𝑉 = 𝑁𝜔 𝐴 × 𝑟 𝑂𝑖
𝑡2
𝑁
𝑀
𝑖
𝑜 𝑑𝑡
𝑡1
𝑖
𝑜
𝑖
= 𝐻2 − 𝐻1 𝑜
𝑁
𝑁
The system starts from rest and we insert the expression for H:
𝑡2
𝑡1
𝑁
𝑀
𝑖
𝑜 𝑑𝑡
= 𝑟 𝑜𝑖 × 𝑚𝑖
𝑁
𝜔 𝐴 × 𝑟 𝑂𝑖
And we need to know the moment
applied
In a similar way that Newtons second law links the kinetics of a problem with the
kinematics, the previous expression does as well.
We need moments though, since this is rotating.
𝑁𝑀 𝑖 𝑜
=?
Well, it’s a Tension T
that is 0.1 meters
out from the radius.
𝑀 = 𝑟 × 𝐹 = −0.1𝑛2 × −𝑇𝑛3 = 0.1𝑇𝑛1
Solving for the moment through time
This is a relatively simple integration: (useful note, 𝑛1 is constant in time)
𝑡2 𝑁 𝑖 𝑜
𝑀 𝑑𝑡
𝑡1
=
𝑡
0.1𝑇𝑛1 𝑑𝑡
0
= 0.1𝑇𝑡 𝑛1
Ok, let’s plug this into the previous equation. (The 4 that magically appears is due to
the fact the spindle has 4 balls).
0.1𝑇𝑡 𝑛1 = 4 𝑟 𝑜𝑖 × 𝑚𝑖
𝑁
𝜔 𝐴 × 𝑟 𝑂𝑖
The usefulness of ‘aligned reference
frames’
The phrase ‘aligned reference frames’ is mine. I designed this system so that 𝑛1 and 𝑎1
are in the same direction. This means that, even if I don’t know the rate of rotation, 𝜔,
I can at least express what it looks like as a vector:
𝑁
𝜔 𝐴 = 𝜔𝑛1 = 𝜔𝑎1
Consider me writing it twice as like a Rosetta stone of my handwriting.
Still filling in some blanks
We need the position vector from O to I :
𝑟 𝑂𝑖 = 0.4𝑎2
Now to calculate the cross product:
𝑁
𝜔 𝐴 × 𝑟 𝑂𝑖 = 𝜔𝑎1 × 0.4𝑎2 = 0.4𝜔𝑎3
0.1𝑇𝑡 𝑛1 = 4 0.4𝑎2 × 𝑚𝑖 0.4𝜔𝑎3
What was the mass again?
The phrase ‘aligned reference frames’ is mine. I designed this system so that 𝑛1 and 𝑎1
are in the same direction. This means that, even if I don’t know the rate of rotation, 𝜔,
I can at least express what it looks like as a vector:
0.1𝑇𝑡 𝑛1 = 12𝑘𝑔 0.4𝑎2 × 0.4𝜔𝑎3
0.1𝑇𝑡 𝑛1 = 12𝑘𝑔 0.16𝜔𝑎1
Man, am I happy I set 𝑛1 = 𝑎1
0.1𝑇𝑡 = 12𝑘𝑔 0.16𝜔
Another quick snippet from my
handwritten notes…
And there you have it. This conversion from rev/min to radians per second is useful!
Lecture 10 Example 2
To pass inspection, steel balls designed for use in
ball bearings must clear the fixed bar A at the top
of their rebound when dropped from rest through
the vertical distance H = 36 in. onto the heavy
inclined steel plate. If the balls which have a
coefficient of restitution of less than 0.7 with the
rebound plate are to be rejected, determine the
position of bar the by specifying h and s. Neglect
any friction during the impact.
Let’s draw some pictures
𝑛
I know that the ‘gravity’ reference frame will
be useful. (again, I made up this phrase.)
𝒋
But there is another frame we need, because
a collision is involved
𝑡
𝒊
Step 1 : Speed at collision
𝑛
If one of the two things colliding ‘can’t move’,
then 𝑡 is aligned with the surface of contact. If
this plate could move, then aim for the center.
Also, it looks like this is pretty much on center.
You know, just having the magnitude of velocity
would be useful. Using geometry, we could find
the normal and tangential components of
velocity.
Important assumption : Restoring force and
deformation force are much greater than gravity
𝑡
Choose your adventure!
I will set and run a Poll:
Conservation of Energy
Kinematics Equations
Conservation of Energy
𝑏𝑎𝑙𝑙
𝑊12
= ∆𝐾𝐸 + ∆𝑃𝐸 = 0
because no energy enters or leaves the system
0 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖 + 𝑃𝐸𝑓 − 𝑃𝐸𝑖
Discussion : Simplifying
1
𝐾𝐸𝑓 = 𝑃𝐸𝑖 → 𝑚𝑣 2 = 𝑚ℎ𝑔
2
Solving for velocity
Now, we have a formula, lets solve for velocity and find a magnitude
𝑣 2 = 2𝑔ℎ
𝑣=
2𝑔ℎ =
2 32.2
Exit
𝑓𝑡
𝑠𝑒𝑐 2
36 𝑖𝑛𝑐ℎ𝑒𝑠
1 𝑓𝑡
𝑓𝑡
= 13.90
12 𝑖𝑛𝑐ℎ𝑒𝑠
𝑠𝑒𝑐
Kinematics
Constant Acceleration means this is true:
𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑥
𝑣𝑓2 = 2𝑔ℎ
𝑣𝑓 =
2𝑔ℎ =
Exit
2 32.2
𝑓𝑡
𝑠𝑒𝑐 2
36 𝑖𝑛𝑐ℎ𝑒𝑠
1 𝑓𝑡
𝑓𝑡
= 13.90
12 𝑖𝑛𝑐ℎ𝑒𝑠
𝑠𝑒𝑐
Solving for normal and tangent velocity
𝑛
𝑗
𝜃 = 10°
𝑗 = − sin 10° 𝑡 + cos 10° 𝑛
𝜃
𝑵𝑻 𝑏𝑎𝑙𝑙
𝑉2
𝑡
= −13.9 − sin 10° 𝑡 + −13.9 cos 10° 𝑛
Applying efficiency.
Ok, so it is time to use the efficiency equation. However, let’s be careful about our
‘time steps’, or the different points in time we are using for labels.
1. Time 1 is when the ball starts to fall
2. Time 2 is just before the ball collides with the plate.
3. Time 3 is just after the ball leaves the plate.
We know the velocity of the ball normal and tangent to the plate at time 2. So,
lets work off time 3 as the ‘after’
𝑒=
𝐵
𝑣3𝑛
𝐴
𝑣2𝑛
−
−
𝐴
𝑣3𝑛
𝐵
𝑣2𝑛
Modeling the plate.
Discussion : Is it fair to say the plate doesn’t move?
𝑒=
𝑏𝑎𝑙𝑙
−𝑣3𝑛
𝑏𝑎𝑙𝑙
𝑣2𝑛
𝑏𝑎𝑙𝑙
𝑣3𝑛
→ 0.7 =
𝑓𝑡
= 9.58
𝑠𝑒𝑐
𝑏𝑎𝑙𝑙
−𝑣3𝑛
−13.9 cos 10°
Velocity as a vector at time 3
𝑵𝑻𝑉 𝑏𝑎𝑙𝑙
3
= −13.9 − sin 10° 𝑡 + 0.7 −13.9 cos 10° 𝑛
𝑵𝑻𝑉 𝑏𝑎𝑙𝑙
3
𝑓𝑡
= 2.41𝑡 + 9.58𝑛
𝑠𝑒𝑐
So, who is up for a trip back to the N Basis?
Back to N
The acceleration due to gravity is in the N basis, in 𝑖 − 𝑗. We
should work in that system.
𝑡 = cos 10° 𝑖 − sin 10° 𝑗
𝑛 = sin 10° 𝑖 + cos 10° 𝑗
𝑵𝑉 𝑏𝑎𝑙𝑙
3
𝑓𝑡
= 4.04𝑖 + 9.02𝑗
𝑠𝑒𝑐
Then, in words, you find the max height a ball would reach, and use the time to get there
to find the value for s. The answers are s = 1.13 ft and h = 1.26 ft, but since this is a
problem we have done so many times, I am going to skip over it.